Pandas dataframe groupby 计算总体标准差
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Pandas dataframe groupby to calculate population standard deviation
提问by neelshiv
I am trying to use groupby and np.std to calculate a standard deviation, but it seems to be calculating a sample standard deviation (with a degrees of freedom equal to 1).
我正在尝试使用 groupby 和 np.std 来计算标准偏差，但它似乎正在计算样本标准偏差（自由度等于 1）。
Here is a sample.
这是一个示例。
#create dataframe
>>> df = pd.DataFrame({'A':[1,1,2,2],'B':[1,2,1,2],'values':np.arange(10,30,5)})
>>> df
A B values
0 1 1 10
1 1 2 15
2 2 1 20
3 2 2 25
#calculate standard deviation using groupby
>>> df.groupby('A').agg(np.std)
B values
A
1 0.707107 3.535534
2 0.707107 3.535534
#Calculate using numpy (np.std)
>>> np.std([10,15],ddof=0)
2.5
>>> np.std([10,15],ddof=1)
3.5355339059327378
Is there a way to use the population std calculation (ddof=0) with the groupby statement? The records I am using are not (not the example table above) are not samples, so I am only interested in population std deviations.
有没有办法在 groupby 语句中使用人口标准计算（ddof=0）？我使用的记录不是（不是上面的示例表）不是样本，所以我只对总体标准偏差感兴趣。
回答by EdChum
You can pass additional args to np.stdin the aggfunction:
您可以np.std在agg函数中传递额外的参数：
In [202]:
df.groupby('A').agg(np.std, ddof=0)
Out[202]:
B values
A
1 0.5 2.5
2 0.5 2.5
In [203]:
df.groupby('A').agg(np.std, ddof=1)
Out[203]:
B values
A
1 0.707107 3.535534
2 0.707107 3.535534
回答by Giorgos Myrianthous
For degree of freedom = 0
为了 degree of freedom = 0
(This means that bins with one number will end up with std=0instead of NaN)
（这意味着带有一个数字的垃圾箱将以std=0代替NaN）
import numpy as np
def std(x):
return np.std(x)
df.groupby('A').agg(['mean', 'max', std])
